Series RLC Circuit and Maximum Current

In a series RLC circuit, the current depends on the frequency of the AC source. The opposition to the current flow in the circuit is called impedance ($Z$). The impedance is determined by the resistance ($R$), inductive reactance ($X_L$), and capacitive reactance ($X_C$). The formula for impedance in a series RLC circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2}$. The current ($I$) in the circuit is given by $I = \frac{V}{Z}$, where $V$ is the source voltage.

Achieving Maximum Current via Resonance

To achieve the maximum current in a series RLC circuit, the impedance ($Z$) must be minimum. This occurs when the inductive reactance ($X_L$) equals the capacitive reactance ($X_C$). This condition is known as resonance. At resonance, the impedance formula simplifies to $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + (0)^2} = R$. Since the resistance ($R$) is constant, the minimum impedance is equal to the resistance, leading to the maximum possible current ($I_{max} = \frac{V}{R}$).

Calculating Capacitance for Resonance

The condition for resonance is $X_L = X_C$. We know that inductive reactance is $X_L = 2\pi f L$ and capacitive reactance is $X_C = \frac{1}{2\pi f C}$, where $f$ is the frequency, $L$ is the inductance, and $C$ is the capacitance.

Setting $X_L = X_C$ gives:

The frequency at which this occurs is the resonant frequency, often denoted as $f_0$. The formula relating resonant frequency, inductance, and capacitance is:

We are given:

• Inductance, $L = 500 \, \text{mH} = 500 \times 10^{-3} \, \text{H} = 0.5 \, \text{H}
• Frequency, $f = 0.4 \, \text{kHz} = 0.4 \times 10^3 \, \text{Hz} = 400 \, \text{Hz}

Since the question asks for the capacitance value to get maximum current, the given frequency of 0.4 kHz must be the resonant frequency ($f_0$).

Step-by-Step Calculation

1. Rearrange the resonant frequency formula to solve for Capacitance ($C$):

   Starting with $f_0 = \frac{1}{2\pi\sqrt{LC}}$, we can isolate $C$:

   $ \sqrt{LC} = \frac{1}{2\pi f_0} $ $ LC = \left(\frac{1}{2\pi f_0}\right)^2 $ $ C = \frac{1}{(2\pi f_0)^2 L} $
2. Substitute the given values into the formula:

   Using $f_0 = 400 \, \text{Hz}$ and $L = 0.5 \, \text{H}$:

   $ C = \frac{1}{(2\pi \times 400)^2 \times 0.5} $
3. Calculate the value of $C$: $ C = \frac{1}{(800\pi)^2 \times 0.5} $ $ C = \frac{1}{(640000 \pi^2) \times 0.5} $ $ C = \frac{1}{320000 \pi^2} $

   Using $\pi \approx 3.14159$, $\pi^2 \approx 9.8696$:

   $ C = \frac{1}{320000 \times 9.8696} $ $ C = \frac{1}{3158272} $ $ C \approx 3.166 \times 10^{-7} \, \text{F} $
4. Convert the result to microfarads ($\mu$F):

   Since $1 \, \text{F} = 10^6 \, \mu\text{F}$:

   $ C \approx 3.166 \times 10^{-7} \times 10^6 \, \mu\text{F} $ $ C \approx 0.3166 \, \mu\text{F} $

Comparing this calculated value with the given options, we find that $0.3166 \, \mu\text{F}$ is approximately equal to $0.32 \, \mu\text{F}$.

Conclusion

Therefore, the value of the capacitance required to obtain maximum current in the series RLC circuit at a frequency of 0.4 kHz is approximately $0.32 \, \mu\text{F}$.